Invariance of curvature under a conformal mapping

Question

Let $\Omega_{1}, \Omega_{2} \subseteq \mathbb{C}$ be bounded domains. Let $\rho$ be a metric on $\Omega_2$ and $h: \Omega_1 \rightarrow \Omega_2$ a conformal mapping. Let $$h^*\rho(z) = \rho(h(z))\cdot|h'(z)| $$ be the pullback metric on $\Omega_1$. According to the definition I have been given, the curvature on $(\Omega_2,\rho)$ is

$$K_{\Omega_2,\rho}(z) = -\frac{\Delta \log(\rho(z))}{\rho(z)^2}$$

It is then claimed that $K_{\Omega_1,h^*\rho}(z) = K_{\Omega_2,\rho}(h(z))$, i.e.

$$ -\frac{\Delta \log(\rho(h(z))\cdot|h'(z)|) }{(\rho(h(z))\cdot|h'(z)|)^2} = -\frac{\Delta \log(\rho(h(z)))}{\rho(h(z))^2}$$

I cannot derive this result however, and quite frankly I find it dubious. Is it true?

Answer 1

1. We can get rid of $|h'|$ under the logarithm, since $\log |h'|$ is harmonic.

2. The composition of a function ($\log \rho$ in this case) with a conformal map $h$ multiplies the Laplacian by $|h'|^2$. This is shown in one line here.

3. By definition of pullback, the denominator is multiplied by $|h'|^2$ too. So $|h'|^2$ cancels, and we are done.

Answer 2

The calculation is fairly painful. I think that Theorem 2.5.4 of this should do the trick.